Pretty sure it is impossible if everything is finite, $|G_i|$ must be a prime or a power of a prime, then either LHS and RHS have different group sizes or they have different number of elements of a particular order. I just don't know about the infinity case, feels like it might not be true for some special case. Need help please.
This is answered in Theorem 90.1 of Infinite Abelian Groups (Vol II) by L. Fuchs, which shows that, for any $n\geq 2$, there is a finite rank abelian group which is the direct sum of two indecomposable groups and of $n$ indecomposable groups.
For the case $n=3$, one example that can be extracted from the general theorem is as follows. All the groups will be subgroups of $\mathbb{Q}^4$.
$G_1$ is generated by the elements $\left(\frac{2}{5^s},0,\frac{1}{5^s},0\right)$, $\left(0,\frac{2}{7^t},0,\frac{1}{7^t}\right)$ for all $s,t>0$ and $\left(0,0,\frac{1}{2},\frac{1}{2}\right)$.
$G_2$ is generated by the elements $\left(\frac{3}{5^s},0,\frac{1}{5^s},0\right)$, $\left(0,\frac{3}{7^t},0,\frac{1}{7^t}\right)$ for all $s,t>0$ and $\left(0,0,\frac{1}{3},\frac{1}{3}\right)$.
$G_3$ is generated by the elements $\left(\frac{1}{5^s},0,0,0\right)$ for all $s>0$.
$G_4$ is generated by the elements $\left(0,\frac{1}{7^t},0,0\right)$ for all $t>0$.
$G_5$ is generated by the elements $\left(0,0,\frac{1}{5^s},0\right)$, $\left(0,0,0,\frac{1}{7^t}\right)$ for all $s,t>0$ and $\left(0,0,\frac{1}{6},\frac{1}{6}\right)$.
It's straightforward to check that $G_1\oplus G_2=G_3\oplus G_4\oplus G_5$. Fuchs quotes a theorem from earlier in the book to prove indecomposability in the more general statement that he gives, but it's probably not too hard to check directly in this specific example.